Transcript module 8b

Principle of the process
Metal
forming
Structure
Process modeling
Rolling
Defects
Design For Manufacturing (DFM)
Process variation
Module 8
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Rolling Process: Mechanics Analysis
• Two opposite rolls and a piece of material flows between
them. The shape of rolls can be designed in a different form
to construct a product with different cross sections.
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Objectives of mechanics analysis
a. Physical
phenomenon
b. Torque
c. Power
d. Productivity
System
parameters
Operating
parameters
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Physical phenomenon
Draft , d  t0  t f
w
L
v0  v f
t
Spreading: Volume before
rolling = the volume after rolling
t 0 w0 L0  t f w f L f
Volume flow rate conservation
t 0 w0 v0  t f w f v f
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Vo < Vr < Vf
Slipping
No-slip point
Work velocity = Roll velocity
Slipping
Forward  Slip  (v f  vr ) / vr
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For the rolling process, the true strain is:
t0
  ln
tf
The average flow stress is the same expression, i.e.
k
Yf 
1+ n
n
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It is the friction between the work and the roll that
drives the workflow between two rolls.
Greater
Friction Force
No-slip point
Lesser
Friction Force
The friction force is developed based on
(1) coefficient of the friction and
(2) compression force of rolls
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Condition to roll- Coefficient of the friction
draft, d = |tf-t0|: dmax
Max.
Possible
Draft
d max   R
2
Radius of the roll
Friction coefficient
Friction causes Rolling
If Friction=0, then draft=0, means NO ROLLING
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Condition to roll- Power to drive the roll and work piece
Roll Force (F) Integrating
“unit roll pressure” over
roll work “contact area”
L
F  w pdL
0
The pressure varies along the
contact length
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L
Contact length
R (t 0  t f )
d
Contact force
L
F  w pdL
F  Y f wL
0
Torque
T  0.5FL
Power
P  2NFL
N: rotation
speed of the roll,
rev / min
Power is a function of D. Increase of D leads to increase of P
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Condition to roll- Power to drive the roll and work piece
When the required power (d) is greater than the supplied
power, the rolling of a work piece with d is not possible.
Therefore, the required power = supplied power will lead to
a critical draft d or maximum d.
Criterion 1:
d max   2 R
Criterion 2:
required power = supplied power 
d max
The actual maximum draft for a rolling system is the smaller
one computed from the two criterions above.
Example:
A 10-in. –wide, 1.0-in – thick plate is to be reduced in a
single pass in a two-high rolling mill to a thickness of 0.8
in. The roll has a radius = 20 in., and its speed = 50
ft/min. The work material has a strength coefficient =
35.000 lb/in.2 and a strain hardening exponent = 0.2.
Determine (a) roll force, (b) roll torque, and (c) power
required to accomplish this operation.
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Given: rolling, t0=1.0 in., tf=0.80 in., w=10.0 in., R=20
in., vr=50 ft/min, flow curve n=0.20 and K=35,000 lb/in2.
Find: (a) F, (b) T, (c) HP.
Draft d=1.0-0.8=0.2 in.
Contact length L = (20×0.20)0.5 = 2.0 in.
True strain ε = ln (1.0/0.8) = ln 1.25= 0.223
Average flow stress
Yf = 35,000(0.223)0.20/1.20 = 21,607 lb/in2
Rolling force F = 21,607(10)(2) = 423,149 lb
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Torque T = 0.5(432,149)(2.0) = 432,149 in-lb.
N = (50 ft/min)/(2π×20/12) = 4.77 rev/min.
Power P=2 π (4.77)(432,149)(2) = 25,929,940 in-lb/min
HP = (25,929,940 in-lb/min)/(396,000) = 65.5 hp
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